Find $\int \dfrac{1}{\sqrt{-x^2-14x-48}}\,dx$. Choose 1 answer: Choose 1 answer: (Choice A) A $\text{arctan}\left(x+7\right)+C$ (Choice B) B $\text{arcsin}\left(x+7\right)+C$ (Choice C) C $\dfrac12\text{arcsin}\left(\dfrac{x+7}{2}\right)+C$ (Choice D) D $\dfrac12\text{arctan}\left(\dfrac{x+7}{2}\right)+C$
Explanation: The integrand is in the form $\dfrac{1}{\sqrt{p(x)}}$ where $p(x)$ is a quadratic expression. This suggests that we should rewrite $p(x)$ by completing the square. Specifically, we will rewrite $p(x)$ as $ k^2-(x+ h)^2$. Then, we will be able to integrate using our knowledge of the derivative of the inverse sine function: $\int \dfrac{1}{\sqrt{ k^2-x^2}}\,dx=\text{arcsin}\left(\dfrac{x}{ k}\right)+C$ [Why is this formula true?] By setting $u=x+ h$ and using $u$ -substitution, we get the following formula: $\int \dfrac{1}{\sqrt{ k^2-(x+ h)^2}}\,dx=\text{arcsin}\left(\dfrac{x+ h}{ k}\right)+C$ We start by rewriting $p(x)$ as $ k^2-(x+ h)^2$ : $\begin{aligned} -x^2-14x-48&=-48-(x^2+14x) \\\\ &=-48+49-(x^2+14x+49) \\\\ &=1-(x+7)^2 \\\\ &={1}^2-(x+{7})^2 \end{aligned}$ Now we can find the integral: $\begin{aligned} &\phantom{=}\int \dfrac{1}{\sqrt{-x^2-14x-48}}\,dx \\\\ &=\int\dfrac{1}{\sqrt{{1}^2-(x+{7})^2}}\,dx \\\\ &=\text{arcsin}\left(\dfrac{x+{7}}{{1}}\right)+C \\\\ &=\text{arcsin}\left(x+7\right)+C \end{aligned}$ In conclusion, $\int \dfrac{1}{\sqrt{-x^2-14x-48}}\,dx=\text{arcsin}\left(x+7\right)+C$